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Comparing proportions involves determining whether there is a significant difference between the percentages of two groups in a given population.
For instance, in our exercise, we're interested in the proportion of Americans satisfied with the quality of the environment in 2016 versus 2018.
This type of comparison is often used in surveys or opinion polls to see if attitudinal shifts have occurred.
To start, we calculate each year's proportion by dividing the number of satisfied individuals by the total respondents for that year.

• In 2016, the satisfied proportion was calculated as \(p_1 = \frac{543}{543 + 440} = 0.55\).
• In 2018, it was \(p_2 = \frac{461}{461 + 532} = 0.46\).

This shows us that fewer Americans were satisfied in 2018 compared to 2016. However, to determine if this difference is statistically significant, further analysis is required.

The significance level, often denoted by \(\alpha\), is a threshold set by the researcher for determining when to reject a null hypothesis.
It reflects how confident we are about the results being reliable and not due to random chance.
The most common significance levels are \(0.05\), \(0.01\), and \(0.10\).In our scenario, the significance level provided is \(0.05\).
This means we're accepting a 5% risk of concluding a difference when there is none.
If our calculated p-value is less than \(0.05\), we reject the null hypothesis, indicating a statistically significant change in the proportion of satisfied Americans from 2016 to 2018.
However, if the p-value is higher than \(0.05\), then we would not have strong enough evidence to claim a significant difference.
Setting this level beforehand prevents bias in interpreting results.

Calculating the p-value is crucial for hypothesis testing as it helps quantify the evidence against the null hypothesis.
The p-value indicates the probability of observing the data, or something more extreme, if the null hypothesis is true.In our exercise, the test statistic \(z\) is already calculated as \(z = 4.09\).
We use this to find the p-value associated with a one-tailed z-test, since we are checking for a decrease.
As calculated, the p-value for \(z = 4.09\) is approximately \(0.00002\).This incredibly small p-value is far less than our predetermined significance level of \(0.05\).
Thus, it provides very strong evidence against the null hypothesis, supporting the conclusion that satisfaction has indeed decreased between 2016 and 2018.
A smaller p-value indicates stronger evidence in favor of the alternative hypothesis.

Z-tests are statistical tests used to determine if there is a significant difference between sample and population means or sample means among groups.
They are particularly useful when the sample size is large (typically \(n \geq 30\)).
Here, we want to spot if there's a drop in satisfaction with environmental quality over time.The z-test involves calculating a z-score, which is the number of standard deviations a data point is from the mean.
It involves steps like determining the pooled proportion, calculating the standard error (SE), and then using these to determine the z-score.For our comparison between 2016 and 2018, the test statistic \(z\) was calculated using:

• Difference in sample proportions: \(p_1 - p_2\)
• Standard Error: \(SE = \sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\)

This calculation yielded a z-score of \(4.09\), which indicates a notably significant difference (well beyond typical thresholds).
A high absolute z-score coupled with a small p-value allows us to confidently conclude a decrease in environmental satisfaction.